If the graph of y = |x| is translated so that the point (1, 1) is moved to (4, 1), what is the equation of the new graph?

1 answer:

4 0

First you graph all of these answers in a calculator:

y= Ix-3I is correct, it moved to (4,1).

y=IxI-3 is incorrect because it moved to (1,-2)

y=Ix +3I is incorrect because it moved to (-2,1)

y=IxI+3 is incorrect because it moved to (1,4)

So your answer is A: y= Ix-3I

You might be interested in

If sin x=cos y,then x+y=45,write true or false

taurus [48]

False........................

3 0

3 years ago

Read 2 more answers

Which answer correct estimates the product of 3.09 × 304.87

miv72 [106K]

The answer is 942.05. You get it by multiplying the two numbers together which should give you the product of 942.0483, since the 0 is not the number 5 or more ( which is the rule to round to a higher number) then you go to the hundredths place value which will be underlined for you... 942.0<u>4</u>83 and look to the right and since 8 is higher than 5, you change the four to a five, thus giving you your answer 942.05

6 0

3 years ago

Find the area of a floor which is covered by 1050 tiles.each tiles measuring 20 square meters

RoseWind [281]

Answer:

area of floor is

1050*20=21000

Step-by-step explanation:

mark me as brainliest

7 0

3 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

Oksana_A [137]

Answer:

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

${n \choose 5 } = \frac{n!}{5!(n-5)!}$

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.